JEE Main · 2005 · 08 May 2005 · Offline · Matrices

We are told that $A^2-A+I=0$. Since this can be written as $A(A-I)=-I$, the matrix $A$ is invertible, so we may multiply the given relation on the left by $A^{-1}$. Using $A^{-1}{A}^2=A$, $A^{-1}A=I$ and $A^{-1}I=A^{-1}$, this becomes $A-I+A^{-1}=0$. Rearranging, the inverse of $A$ is $A^{-1}=I-A$.